Add two numbers represented by linked lists

Given two numbers represented by two lists, write a function that returns sum list. The sum list is list representation of addition of two input numbers.

Example 1

Input:
  First List: 5->6->3  // represents number 365
  Second List: 8->4->2 //  represents number 248
Output
  Resultant list: 3->1->6  // represents number 613

Example 2

Input:
  First List: 7->5->9->4->6  // represents number 64957
  Second List: 8->4 //  represents number 48
Output
  Resultant list: 5->0->0->5->6  // represents number 65005

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Solution
Traverse both lists. One by one pick nodes of both lists and add the values. If sum is more than 10 then make carry as 1 and reduce sum. If one list has more elements than the other then consider remaining values of this list as 0. Following is the implementation of this approach.

C++

// C++ program to add two numbers  
// represented by linked list 
#include <bits/stdc++.h> 
using namespace std; 
  
/* Linked list node */
class Node  
{  
    public: 
    int data;  
    Node* next;  
};  
  
/* Function to create a new node with given data */
Node *newNode(int data)  
{  
    Node *new_node = new Node(); 
    new_node->data = data;  
    new_node->next = NULL;  
    return new_node;  
}  
  
/* Function to insert a node at the 
beginning of the Singly Linked List */
void push(Node** head_ref, int new_data)  
{  
    /* allocate node */
    Node* new_node = newNode(new_data);  
  
    /* link the old list off the new node */
    new_node->next = (*head_ref);  
  
    /* move the head to point to the new node */
    (*head_ref) = new_node;  
}  
  
/* Adds contents of two linked lists and  
return the head node of resultant list */
Node* addTwoLists (Node* first, Node* second)  
{  
    Node* res = NULL; // res is head node of the resultant list  
    Node *temp, *prev = NULL;  
    int carry = 0, sum;  
  
    while (first != NULL || second != NULL) //while both lists exist  
    {  
        // Calculate value of next digit in resultant list.  
        // The next digit is sum of following things  
        // (i) Carry  
        // (ii) Next digit of first list (if there is a next digit)  
        // (ii) Next digit of second list (if there is a next digit)  
        sum = carry + (first? first->data: 0) + 
                      (second? second->data: 0);  
  
        // update carry for next calulation  
        carry = (sum >= 10)? 1 : 0;  
  
        // update sum if it is greater than 10  
        sum = sum % 10;  
  
        // Create a new node with sum as data  
        temp = newNode(sum);  
  
        // if this is the first node then 
        // set it as head of the resultant list  
        if(res == NULL)  
            res = temp;  
              
        // If this is not the first  
        // node then connect it to the rest. 
        else  
            prev->next = temp;  
  
        // Set prev for next insertion  
        prev = temp;  
  
        // Move first and second pointers to next nodes  
        if (first) first = first->next;  
        if (second) second = second->next;  
    }  
  
    if (carry > 0)  
    temp->next = newNode(carry);  
  
    // return head of the resultant list  
    return res;  
}  
  
// A utility function to print a linked list  
void printList(Node *node)  
{  
    while(node != NULL)  
    {  
        cout << node->data << " ";  
        node = node->next;  
    }  
    cout<<endl;  
}  
  
/* Driver code */
int main(void)  
{  
    Node* res = NULL;  
    Node* first = NULL;  
    Node* second = NULL;  
  
    // create first list 7->5->9->4->6  
    push(&first, 6);  
    push(&first, 4);  
    push(&first, 9);  
    push(&first, 5);  
    push(&first, 7);  
    printf("First List is ");  
    printList(first);  
  
    // create second list 8->4  
    push(&second, 4);  
    push(&second, 8);  
    cout<<"Second List is ";  
    printList(second);  
  
    // Add the two lists and see result  
    res = addTwoLists(first, second);  
    cout<<"Resultant list is ";  
    printList(res);  
  
return 0;  
}  
  
// This code is contributed by rathbhupendra 

C

#include<stdio.h> 
#include<stdlib.h> 
  
/* Linked list node */
struct Node 
{ 
    int data; 
    struct Node* next; 
}; 
  
/* Function to create a new node with given data */
struct Node *newNode(int data) 
{ 
    struct Node *new_node = (struct Node *) malloc(sizeof(struct Node)); 
    new_node->data = data; 
    new_node->next = NULL; 
    return new_node; 
} 
  
/* Function to insert a node at the beginning of the Singly Linked List */
void push(struct Node** head_ref, int new_data) 
{ 
    /* allocate node */
    struct Node* new_node = newNode(new_data); 
  
    /* link the old list off the new node */
    new_node->next = (*head_ref); 
  
    /* move the head to point to the new node */
    (*head_ref)    = new_node; 
} 
  
/* Adds contents of two linked lists and return the head node of resultant list */
struct Node* addTwoLists (struct Node* first, struct Node* second) 
{ 
    struct Node* res = NULL; // res is head node of the resultant list 
    struct Node *temp, *prev = NULL; 
    int carry = 0, sum; 
  
    while (first != NULL || second != NULL) //while both lists exist 
    { 
        // Calculate value of next digit in resultant list. 
        // The next digit is sum of following things 
        // (i)  Carry 
        // (ii) Next digit of first list (if there is a next digit) 
        // (ii) Next digit of second list (if there is a next digit) 
        sum = carry + (first? first->data: 0) + (second? second->data: 0); 
  
        // update carry for next calulation 
        carry = (sum >= 10)? 1 : 0; 
  
        // update sum if it is greater than 10 
        sum = sum % 10; 
  
        // Create a new node with sum as data 
        temp = newNode(sum); 
  
        // if this is the first node then set it as head of the resultant list 
        if(res == NULL) 
            res = temp; 
        else // If this is not the first node then connect it to the rest. 
            prev->next = temp; 
  
        // Set prev for next insertion 
        prev  = temp; 
  
        // Move first and second pointers to next nodes 
        if (first) first = first->next; 
        if (second) second = second->next; 
    } 
  
    if (carry > 0) 
      temp->next = newNode(carry); 
  
    // return head of the resultant list 
    return res; 
} 
  
// A utility function to print a linked list 
void printList(struct Node *node) 
{ 
    while(node != NULL) 
    { 
        printf("%d ", node->data); 
        node = node->next; 
    } 
    printf("\n"); 
} 
  
/* Driver program to test above function */
int main(void) 
{ 
    struct Node* res = NULL; 
    struct Node* first = NULL; 
    struct Node* second = NULL; 
  
    // create first list 7->5->9->4->6 
    push(&first, 6); 
    push(&first, 4); 
    push(&first, 9); 
    push(&first, 5); 
    push(&first, 7); 
    printf("First List is "); 
    printList(first); 
  
    // create second list 8->4 
    push(&second, 4); 
    push(&second, 8); 
    printf("Second List is "); 
    printList(second); 
  
    // Add the two lists and see result 
    res = addTwoLists(first, second); 
    printf("Resultant list is "); 
    printList(res); 
  
   return 0; 
} 

Java

// Java program to add two numbers represented by linked list 
  
class LinkedList { 
  
    static Node head1, head2; 
  
    static class Node { 
  
        int data; 
        Node next; 
  
        Node(int d) { 
            data = d; 
            next = null; 
        } 
    } 
  
    /* Adds contents of two linked lists and return the head node of resultant list */
    Node addTwoLists(Node first, Node second) { 
        Node res = null; // res is head node of the resultant list 
        Node prev = null; 
        Node temp = null; 
        int carry = 0, sum; 
  
        while (first != null || second != null) //while both lists exist 
        { 
            // Calculate value of next digit in resultant list. 
            // The next digit is sum of following things 
            // (i)  Carry 
            // (ii) Next digit of first list (if there is a next digit) 
            // (ii) Next digit of second list (if there is a next digit) 
            sum = carry + (first != null ? first.data : 0) 
                    + (second != null ? second.data : 0); 
  
            // update carry for next calulation 
            carry = (sum >= 10) ? 1 : 0; 
  
            // update sum if it is greater than 10 
            sum = sum % 10; 
  
            // Create a new node with sum as data 
            temp = new Node(sum); 
  
            // if this is the first node then set it as head of 
            // the resultant list 
            if (res == null) { 
                res = temp; 
            } else // If this is not the first node then connect it to the rest. 
            { 
                prev.next = temp; 
            } 
  
            // Set prev for next insertion 
            prev = temp; 
  
            // Move first and second pointers to next nodes 
            if (first != null) { 
                first = first.next; 
            } 
            if (second != null) { 
                second = second.next; 
            } 
        } 
  
        if (carry > 0) { 
            temp.next = new Node(carry); 
        } 
  
        // return head of the resultant list 
        return res; 
    } 
    /* Utility function to print a linked list */
  
    void printList(Node head) { 
        while (head != null) { 
            System.out.print(head.data + " "); 
            head = head.next; 
        } 
        System.out.println(""); 
    } 
  
    public static void main(String[] args) { 
        LinkedList list = new LinkedList(); 
  
        // creating first list 
        list.head1 = new Node(7); 
        list.head1.next = new Node(5); 
        list.head1.next.next = new Node(9); 
        list.head1.next.next.next = new Node(4); 
        list.head1.next.next.next.next = new Node(6); 
        System.out.print("First List is "); 
        list.printList(head1); 
  
        // creating seconnd list 
        list.head2 = new Node(8); 
        list.head2.next = new Node(4); 
        System.out.print("Second List is "); 
        list.printList(head2); 
  
        // add the two lists and see the result 
        Node rs = list.addTwoLists(head1, head2); 
        System.out.print("Resultant List is "); 
        list.printList(rs); 
    } 
} 
  
// this code has been contributed by Mayank Jaiswal 

Python

# Python program to add two numbers represented by linked list 
  
# Node class 
class Node: 
  
    # Constructor to initialize the node object 
    def __init__(self, data): 
        self.data = data 
        self.next = None
  
class LinkedList: 
  
    # Function to initialize head 
    def __init__(self): 
        self.head = None
  
    # Function to insert a new node at the beginning 
    def push(self, new_data): 
        new_node = Node(new_data) 
        new_node.next = self.head 
        self.head = new_node 
  
    # Add contents of two linked lists and return the head 
    # node of resultant list 
    def addTwoLists(self, first, second): 
        prev = None
        temp = None
        carry = 0 
  
        # While both list exists 
        while(first is not None or second is not None): 
  
            # Calculate the value of next digit in 
            # resultant list 
            # The next digit is sum of following things 
            # (i) Carry 
            # (ii) Next digit of first list (if ther is a 
            # next digit) 
            # (iii) Next digit of second list ( if there 
            # is a next digit) 
            fdata = 0 if first is None else first.data 
            sdata = 0 if second is None else second.data 
            Sum = carry + fdata + sdata 
  
            # update carry for next calculation 
            carry = 1 if Sum >= 10 else 0
  
            # update sum if it is greater than 10 
            Sum = Sum if Sum < 10 else Sum % 10
  
            # Create a new node with sum as data 
            temp = Node(Sum) 
  
            # if this is the first node then set it as head 
            # of resultant list 
            if self.head is None: 
                self.head = temp 
            else : 
                prev.next = temp  
  
            # Set prev for next insertion 
            prev = temp 
  
            # Move first and second pointers to next nodes 
            if first is not None: 
                first = first.next
            if second is not None: 
                second = second.next
  
        if carry > 0: 
            temp.next = Node(carry) 
  
    # Utility function to print the linked LinkedList 
    def printList(self): 
        temp = self.head 
        while(temp): 
            print temp.data, 
            temp = temp.next
  
# Driver program to test above function 
first = LinkedList() 
second = LinkedList() 
  
# Create first list 
first.push(6) 
first.push(4) 
first.push(9) 
first.push(5) 
first.push(7) 
print "First List is ", 
first.printList() 
  
# Create second list 
second.push(4) 
second.push(8) 
print "\nSecond List is ", 
second.printList() 
  
# Add the two lists and see result 
res = LinkedList() 
res.addTwoLists(first.head, second.head) 
print "\nResultant list is ", 
res.printList() 
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

C#

// C# program to add two numbers  
// represented by linked list 
using System; 
  
public class LinkedList 
{ 
  
    Node head1, head2; 
  
    public class Node  
    { 
        public int data; 
        public Node next; 
  
        public Node(int d)  
        { 
            data = d; 
            next = null; 
        } 
    } 
  
    /* Adds contents of two linked lists  
    and return the head node of resultant list */
    Node addTwoLists(Node first, Node second) 
    { 
        // res is head node of the resultant list 
        Node res = null;  
        Node prev = null; 
        Node temp = null; 
        int carry = 0, sum; 
  
        // while both lists exist 
        while (first != null || second != null)  
        { 
            // Calculate value of next digit in resultant list. 
            // The next digit is sum of following things 
            // (i) Carry 
            // (ii) Next digit of first list (if there is a next digit) 
            // (ii) Next digit of second list (if there is a next digit) 
            sum = carry + (first != null ? first.data : 0) 
                    + (second != null ? second.data : 0); 
  
            // update carry for next calulation 
            carry = (sum >= 10) ? 1 : 0; 
  
            // update sum if it is greater than 10 
            sum = sum % 10; 
  
            // Create a new node with sum as data 
            temp = new Node(sum); 
  
            // if this is the first node then set it as head of 
            // the resultant list 
            if (res == null)  
            { 
                res = temp; 
            }  
              
            // If this is not the first 
            // node then connect it to the rest. 
            else 
            { 
                prev.next = temp; 
            } 
  
            // Set prev for next insertion 
            prev = temp; 
  
            // Move first and second pointers to next nodes 
            if (first != null) 
            { 
                first = first.next; 
            } 
            if (second != null)  
            { 
                second = second.next; 
            } 
        } 
  
        if (carry > 0) 
        { 
            temp.next = new Node(carry); 
        } 
  
        // return head of the resultant list 
        return res; 
    } 
    /* Utility function to print a linked list */
  
    void printList(Node head)  
    { 
        while (head != null)  
        { 
            Console.Write(head.data + " "); 
            head = head.next; 
        } 
    Console.WriteLine(""); 
    } 
  
    // Driver code 
    public static void Main(String[] args)  
    { 
        LinkedList list = new LinkedList(); 
  
        // creating first list 
        list.head1 = new Node(7); 
        list.head1.next = new Node(5); 
        list.head1.next.next = new Node(9); 
        list.head1.next.next.next = new Node(4); 
        list.head1.next.next.next.next = new Node(6); 
        Console.Write("First List is "); 
        list.printList(list.head1); 
  
        // creating seconnd list 
        list.head2 = new Node(8); 
        list.head2.next = new Node(4); 
        Console.Write("Second List is "); 
        list.printList(list.head2); 
  
        // add the two lists and see the result 
        Node rs = list.addTwoLists(list.head1, list.head2); 
        Console.Write("Resultant List is "); 
        list.printList(rs); 
    } 
} 
  
// This code contributed by Rajput-Ji 

Output:

  First List is 7 5 9 4 6
  Second List is 8 4
  Resultant list is 5 0 0 5 6

Time Complexity: O(m + n) where m and n are number of nodes in first and second lists respectively.

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

My Personal Notes

Add two numbers represented by linked lists | Set 1

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python

C#

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python

C#

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